# Quick Proof on real numbers

• Sep 14th 2011, 06:36 PM
Quick Proof on real numbers
For all positive real numbers x, (xˆ2)-x≥0.

I don't think this is true since for instance x can equal 2/3 and (2/3)ˆ2-(2/3)=-2/9 which is not greater than or equal to 0. Am i wrong?
• Sep 14th 2011, 06:43 PM
Sudharaka
Re: Quick Proof on real numbers
Quote:

For all positive real numbers x, (xˆ2)-x≥0.

I don't think this is true since for instance x can equal 2/3 and (2/3)ˆ2-(2/3)=-2/9 which is not greater than or equal to 0. Am i wrong?

You are correct. When x is less than one the statement is wrong. However,

$x^2-x\geq 0\mbox{ whenever x\geq 1}$ is a correct statement.
• Sep 14th 2011, 06:50 PM
Re: Quick Proof on real numbers
I see, so is the original proof "For all positive real numbers x, (xˆ2)-x≥0. " is only correct if you set the parameter x≥1.
So the original statement is false since 2/3 is a positive real number, and (2/3)ˆ2-(2/3)=-2/9 which is not greater than or equal to 0. Correct? Just want to make sure whats in my head is correct.
• Sep 14th 2011, 08:26 PM
Sudharaka
Re: Quick Proof on real numbers
Quote:

I see, so is the original proof "For all positive real numbers x, (xˆ2)-x≥0. " is only correct if you set the parameter x≥1.
So the original statement is false since 2/3 is a positive real number, and (2/3)ˆ2-(2/3)=-2/9 which is not greater than or equal to 0. Correct? Just want to make sure whats in my head is correct.

Yes. Correct.
• Sep 14th 2011, 08:51 PM
Prove It
Re: Quick Proof on real numbers
Quote:

For all positive real numbers x, (xˆ2)-x≥0.

I don't think this is true since for instance x can equal 2/3 and (2/3)ˆ2-(2/3)=-2/9 which is not greater than or equal to 0. Am i wrong?

If you want to prove this directly, you could complete the square...

\displaystyle \begin{align*} x^2 - x &\geq 0 \\ x^2 - x + \left(-\frac{1}{2}\right)^2 &\geq \left(-\frac{1}{2}\right)^2 \\ \left(x - \frac{1}{2}\right)^2 &> \frac{1}{4} \\ \left|x - \frac{1}{2}\right| &> \frac{1}{2} \\ x - \frac{1}{2} \leq -\frac{1}{2} \textrm{ or }x - \frac{1}{2} &\geq \frac{1}{2} \\ x \leq 0 \textrm{ or } x &\geq 1 \end{align*}

So the inequality $\displaystyle x^2 - x \geq 0$ is only true when $\displaystyle x \leq 0$ or $\displaystyle x \geq 1$.
• Sep 15th 2011, 04:05 AM
HallsofIvy
Re: Quick Proof on real numbers
S0mewhat simpler, I think:

$x^2- x= x(x- 1)\ge 0$ if and only if both factors on the left have the same sigh. If $x\ge 1$, x> 0 and $x-1\ge 0$, so that is true. If $x\le 0$, x< 1 so that is true. But if $0< x< 1$, x is positive while x- 1 is negative so it is not true.